# Problem in proof: Root of any power $n$ of a positive integer is either irrational or integer.

The proof in the book (Elementary Number Theory, by J.V. Uspensky, Heaslet) is given as follows:

Suppose, this root is represented by the irreducible fraction: $\frac{r}{s}$. Also, $A,r,s,n \in \mathbb {Z} \text{ and all } \gt 0$. Need prove $(\frac{r}{s})^n = A$ is either irrational or integer.

Can state as : $r^n = As^n$, whence it follows that $s^n \mid r^n => s \mid r^n$. Consequently, $(r^n,s) = s$, but by the corollary $(r^n, s) =1$ and so $s=1, A = r^n$. Consequently, if $A$ is not an nth power of an integer,it must be an irrational.

My confusion lies in two parts:

(i) how it implies that : $s \mid r^n$.

(ii) The corollary $(r^n, s) =1$ is seemingly derived out of $(r,s) = 1$, but is not clear how.

In fact, I feel this approach so confusing that would like an alternate approach.

(i) $s^n\mid r^n$ means that $r^n$ is some multiple of $s^n$ (in fact, it is $A$ times $s^n$). Also, it is known that $s^n$ is a multiple of $s$ (it is $s^{n-1}$ times $s$).

From this we conclude that $r^n$ is a multiple of $s$ (it is $As^{n-1}$ times $s$). We write this is $s\mid r^n$.

(ii) if $(r^n,s)\neq1$, then $r^n$ and $s$ must have at least one common prime factor, $p$. If $p$ divides $r^n$, then $p$ divides $r$ (this follows from the defining property of a prime: if $p$ is a prime, and divides a product of numbers, then it divides at least one of the factors in that product).

But if $p$ divides both $s$ and $r$, then it divides $(s,r)$, so $(s,r)\neq1$, which is a contradiction. Thus we must have $(s,r^n)=1$.

As for alternate approaches, I don't know that it's going to help, because this proof comes very naturally. The standard way of showing that a number is irrational is to assume it is rational, and derive a contradiction. The defining property of $n$th roots is that you undo them by raising to the power of $n$. And basically the only tool we have to show that equations do not have integer solutions when it has real solutions (which this clearly has) is divisibility, so we need to use that in some way. This proof follows naturally from these facts.

(i) how it implies that : $s \mid r^n$.

It follows from the transitivity of the "divides" relation: $\;s \mid s^n \mid r^n\,$.

(ii) The corollary $(r^n, s) =1$ is seemingly derived out of $(r,s) = 1$, but is not clear how.

One way is from $(a\cdot b, c) \;\mid\; (a,c) \cdot (b,c)\,$, which implies $(r^n,s) \;\mid\; (r,s)^n = 1\,$.

In fact, I feel this approach so confusing that would like an alternate approach.

• Superb reply. Just one thing left to make it a great answer, i.e. an excellent, symbolic, yet crisp proof of: $(a.b,c) \mid (a,c) \cdot (b,c)$. – jiten Dec 11 '17 at 11:18
• @jiten See for example the top answer to Does $\gcd(a,bc)$ divides $\gcd(a, b)\gcd(a, c)$?. – dxiv Dec 11 '17 at 16:23